On the quartic system $a^4+b^4+x_1^4=2y_1^4,\;b^4+c^4+x_2^4=2y_2^4,\;c^4+a^4+x_3^4=2y_3^4$

Question

I. Quadruples

In the previous post about $x^4+y^4+z^4=3t^2$, quite a lot of solutions share common terms such as the "cyclic",

$$\color{blue}{11}^4+\color{blue}{23}^4+11^4\,=\,3u_1^2\\ \,\color{blue}{23}^4+\color{blue}{85}^4+157^4=3u_2^2\\ \color{blue}{85}^4+\color{blue}{47}^4+37^4\,=\,3u_3^2\\ \color{blue}{47}^4+\color{blue}{11}^4+13^4=3u_4^2$$

or another set,

$$\color{blue}{17}^4+\color{blue}{91}^4+593^4 \,=\,3v_1^2\\ \,\color{blue}{91}^4+\color{blue}{115}^4+293^4=3v_2^2\\ \color{blue}{115}^4+\color{blue}{89}^4+41^4\,=\,3v_3^2\\ \color{blue}{89}^4+\color{blue}{17}^4+547^4=3v_4^2$$

Whether this is just a coincidence is still uncertain, but the cubic version of this has an explanation via an infinite family.

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II. Triples and triangles

But I recall from a 2016 post there is a similar quartic system that does have an explanation. Given the triple $(a,b,c)=(195,264,325)$, then,

$$\color{blue}{195}^4+\color{blue}{264}^4+(195+264)^4 = 2\times399^4\\ \color{blue}{264}^4+\color{blue}{325}^4+(264+325)^4 = 2\times511^4\\ \color{blue}{325}^4+\color{blue}{195}^4+(325+195)^4 = 2\times455^4$$

and infinitely many. These triples can be found in the OEIS as,

$a = 195, 264, 765, 13464, 3515, 4641,\dots$ (A351803)

$b = 264, 325, 1064, 27265, 6528, 34200,\dots$ (A351802)

$c = 325, 440, 5016, 39360, 14800, 70720,\dots$ (A351801)

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III. Elliptic curve

Since $a^4+b^4+(a+b)^4=2(a^2+ab+b^2)^2$, to find these $(a,b,c)$, one can solve the simultaneous,

$$a^2+264a+264^2=w_1^2$$ $$a^2+325a+325^2=w_2^2$$

Two quadratics to be made squares is birationally equivalent to an elliptic curve. Since this has a non-trivial rational point $a=195$, then there infinitely many.

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IV. Question

There are also infinitely many quadruples $(a,b,c,d)$ with $0<a<b<c<d$ and $\text{GCD}(a,b,c,d)=1$ which satisfies the cyclic system,

$$a^4+b^4+(a+b)^4=2{y_1}^4\\a^4+c^4+(a+c)^4=2{y_2}^4\\a^4+d^4+(a+d)^4=2{y_3}^4\\b^4+c^4+(b+c)^4\,=\,2{y_4}^4\\b^4+d^4+(b+d)^4=2{y_5}^4\\c^4+d^4+(c+d)^4=2{y_6}^4$$

such that at least five of the $y_k$ are integers. However, using brute-search, what are the smallest? The ones I've found are,

$(a,b,c,d)=(195,264,325,440)\,$ and only $y_3 \neq \text{integer}$.

$(a,b,c,d)=(1785, 7616, 8415, 11704)\,$ and only $y_1 \neq \text{integer}$.

while Tomita found,

$(a,b,c,d)=(17160, 21125, 23232, 28600)\,$ and only $y_4 \neq \text{integer}$.

Note: These quadruples imply a pair of triples that share two common terms. For example, the first implies $(a,b,c)=(195,264,325)\,$ and $\,(a,b,c)=(264,325,440)$.

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V. Update

I realized Tomita's solution above has the form,

$(a,b,c,d)=(264m,\, 325m,\, 264n,\, 325n)$

which utilized the first triple $(a,b,c)=(195,264,325)$. So the system reduces to,

\begin{align} 511^2 m^2 &=y_1^2\\ 264^2 \,(m^2 + m n + n^2) &=y_2^2\\ 264^2 m^2 + 85800 m n + 325^2 n^2 &=y_3^2\\ 325^2 m^2 + 85800 m n + 264^2 n^2 &=y_4^2\\ 325^2 \,(m^2 + m n + n^2) &=y_5^2\\ 511^2 n^2 &= y_6^2 \end{align}

Thus, we need only choose two equations such as,

\begin{align} m^2 + m n + n^2 &=y_2^2\\ 264^2 m^2 + 85800 m n + 325^2 n^2 &=y_3^2\\ \end{align}

with initial solution $(m,n)=(65,88)$. One can then use an elliptic curve to find infinitely more $(m,n)$.

Answer 1

$1.\ (a,b,c,d)=(264m,325m,264n,325n)$

$$ m^2+mn+n^2=u^2 \tag{1}$$
$$264^2m^2 + 85800mn + 325^2n^2=v^2 \tag{2}$$ First, we get a solution of $(1)$ using Eisenstein integer. $$(m,n)=(p^2-q^2, q^2-2pq)$$ where $p$ and $q$ are any inetgers. Next, substitute $(m,n)$ into equation $(2).$ We have $$V^2 = 69696U^4-171600U^3+368908U^2-250900U+89521$$ where $U=\dfrac{p}{q}.$

This equation can be transformed to Weierstrass form : $$E: Y^2=X^3+X^2-1704184580X -10778950922400$$ $E$ has has rank $2$ and generators are $P(X,Y)=(48703, 4663386)$ and $(46890, 3522750).$

We searched the rational points $(U,V)$ with height$(U)<10000$ as follows.

(m,n),    (a,b,c,d)
(65, 88), (17160, 21125, 23232, 28600)
(-2680, -16541), (5375825, 4366824, 871000, 707520)
(167125, 481536), (44121000, 54315625, 127125504, 156499200)
(3540497235, 13784745629), (934691270040, 1150661601375, 3639172846056, 4480042329425)

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$2.\ (a,b,c,d)=(65m,88m,65n,88n)$

$$ m^2+mn+n^2=u^2 \tag{1}$$
$$88^2m^2+(88m)(65n)+65^2n^2=v^2 \tag{2}$$ First, we get a solution of $(1)$ using Eisenstein integer. $$(m,n)=(p^2-q^2, q^2-2pq)$$ where $p$ and $q$ are any inetgers. Next, substitute $(m,n)$ into equation $(2).$ We have $$V^2 = 7744U^4-11440U^3+7132U^2-5460U+6249$$ where $U=\dfrac{p}{q}.$ This equation can be transformed to Weierstrass form : $$E: Y^2=X^3+X^2-9253860X + 101791008$$ $E$ has has rank $2$ and generators are $P(X,Y)=(3336,79800)$ and $(4818,259578).$

We searched the rational points $(U,V)$ with height$(U)<10000$ as follows.

(m,n), (a,b,c,d)
(3, 5), (195, 264, 325, 440)
(107457, 462528), (2328235, 3152072, 10021440, 13567488)
(514605, 4552064), (33449325, 45285240, 295884160, 400581632)
(18362175, 21732273), (397847125, 470865915, 538623800, 637480008)
(21732273, 24859560), (470865915, 538623800, 637480008, 729213760)